In general, for positive integers $ n$ and $ k$, prove that Diophantine equation $ \prod_{i=1}^n x_i = k\sum_{i=1}^n x_i$ has a finite number of solutions made of positive integers. Assume $ 1 \leq x_1 \leq x_2 \leq \cdots \leq x_n$. Then \[ \frac {1} {k} = \sum_{i=1}^{n} \frac {1} {\prod_{j \neq i} x_j} \leq \frac {n} {\prod_{i=1}^{n-1} x_i}, \ \textrm{ hence } \ \prod_{i=1}^{n-1} x_i \leq nk.\] Therefore $ 1 \leq x_i \leq nk$, $ 1 \leq i \leq n-1$. Since $ x_n$ is uniquely determined, the conclusion ensues.

orznorz, is that enough to prove that there are finitely many solutions?

i think two answers above are both right

Let suppose that $a=max(a,b,c)$. We have $abc=2009(a+b+c)<2009*3*a$, so $bc<2009*3$ , which gives finite values of $b,c$.If $bc<2009$ because $a<a+b+c$ , we have $abc<2009(a+b+c)$ , so $bc>2009$. Because $abc=2009(a+b+c)$ , we obtain $a(bc-2009)=2009(b+c)$.So $a=\frac{2009(b+c)}{bc-2009}$.We know that $a$ is a positive integer so we chose only the values for $b,c$ such that $bc-2009 | 2009(b+c) $ but we know that these are finite. So we have a finite value for $a$.